Answer:
0.172 M
Explanation:
The reaction for the first titration is:
First we <u>calculate how many HCl moles reacted</u>, using the <em>given concentration and volume</em>:
- 19.6 mL * 0.189 M = 3.704 mmol HCl
As one HCl mol reacts with one NaOH mol, <em>there are 3.704 NaOH mmoles in 25.0 mL of solution</em>. With that in mind we <u>determine the NaOH solution concentration</u>:
- 3.704 mmol / 25.0 mL = 0.148 M
As for the second titration:
- H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
We <u>determine how many NaOH moles reacted</u>:
- 34.9 mL * 0.148 M = 5.165 mmol NaOH
Then we <u>convert NaOH moles into H₃PO₄ moles</u>, using the <em>stoichiometric coefficients</em>:
- 5.165 mmol NaOH * = 1.722 mmol H₃PO₄
Finally we <u>determine the H₃PO₄ solution concentration</u>:
- 1.722 mmol / 10.0 mL = 0.172 M
Answer:
6.23 x 10^23 molecules
Explanation:
First find the number of moles of BH3 from the information given. We know the amount of grams present and we can find the molar mass which is 13.84.
We know that moles is grams divided by molar mass so we get 14.32/13.84 which is 1.03 moles.
Finally, to figure out the number of molecules, we multiply 1.03 by Avogadro's number which is 6.022x10^23 and we get 6.23x10^23 molecules.
Answer:
3N2Cl4 + 3CH4--> 3CCl4 + 3N2 + 6H2
This simplifies to:
N2Cl4 + CH4 --> CCl4 + N2 + 2H2
Answer: 0.335 moles
Explanation:
Given that,
Amount of moles of KBr solute (n) = ?
Volume of KBr solution (v) = 0.67 L
Concentration of KBr solution (c) = 0.50M
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
make n the subject formula
n = c x v
n = 0.50M x 0.67L
n = 0.335 moles
Thus, there are 0.335 moles of KBr in 0.67 L of a 0.50 M KBr solution
Answer:
19.32 g/cm³
Explanation:
The density will remain the same no matter how many times you cut the gold. The density is g/cm³ or g/mL. Density is essentially how many grams 1 mL of a compound weighs.